Lucky10
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Can anyone help to me with the molar mass and questions 1-3 please
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Lucky10
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S.G.
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(Original post by Lucky10)
Can anyone help to me with the molar mass and questions 1-3 please
Moles=mass/molar mass

Molar mass is just the relative atomic mass of all the atoms in the molecule added together. So if a molecule has say 1 carbon and 4 hydrogens the molar mass is 1(12) + 4(1) = 16.
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Lucky10
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(Original post by SGHD26716)
Moles=mass/molar mass

Molar mass is just the relative atomic mass of all the atoms in the molecule added together. So if a molecule has say 1 carbon and 4 hydrogens the molar mass is 1(12) + 4(1) = 16.
I don't understand how you do that for ones I have?
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S.G.
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(Original post by Lucky10)
I don't understand how you do that for ones I have?
For question 2b, you need to find the moles of propanol burned. You know that 2.67g was burnt. Now all you need is the molar mass of propanol. Propanol has 3 carbons, 8 hydrogen s and 1 oxygen.

So molar mass is 3(12) + 8(1) + 1(16) = 60

So moles is 2.67/60 = 0.0445 moles burnt
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Lucky10
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(Original post by SGHD26716)
For question 2b, you need to find the moles of propanol burned. You know that 2.67g was burnt. Now all you need is the molar mass of propanol. Propanol has 3 carbons, 8 hydrogen s and 1 oxygen.

So molar mass is 3(12) + 8(1) + 1(16) = 60

So moles is 2.67/60 = 0.0445 moles burnt
How would you work out the molar mass of fuel
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S.G.
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(Original post by Lucky10)
How would you work out the molar mass of fuel
In the periodic table you have the elements and each element has two numbers on it. One is the atomic number and one is the relative atomic mass. You are interested in the relative atomic mass.

Carbon has a relative atomic mass of 12
Hydrogen has a relative atomic mass of 1
Oxygen has a relative atomic mass of 16

Propanol has 3 carbons 8 hydrogens and 1 oxygen. You simply add them together the relative atomic masses.

Before doing these questions I think you should read your textbook and notes carefully.

Propanol is the fuel and so is ethanol
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Lucky10
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(Original post by SGHD26716)
In the periodic table you have the elements and each element has two numbers on it. One is the atomic number and one is the relative atomic mass. You are interested in the relative atomic mass.

Carbon has a relative atomic mass of 12
Hydrogen has a relative atomic mass of 1
Oxygen has a relative atomic mass of 16

Propanol has 3 carbons 8 hydrogens and 1 oxygen. You simply add them together the relative atomic masses.

Before doing these questions I think you should read your textbook and notes carefully.

Propanol is the fuel and so is ethanol
Would propanol be 60?
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S.G.
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(Original post by Lucky10)
Would propanol be 60?
Yes
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Lucky10
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(Original post by SGHD26716)
Yes
How would you do 2a and C
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S.G.
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(Original post by Lucky10)
How would you do 2a and C
2a) q=mxcxdeltat
2c) q divided by moles.

Do you have a textbook? If you don't get this one, assuming you do OCR https://www.amazon.co.uk/gp/aw/d/019...nQL&ref=plSrch
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Lucky10
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(Original post by SGHD26716)
2a) q=mxcxdeltat
2c) q divided by moles.

Do you have a textbook? If you don't get this one, assuming you do OCR https://www.amazon.co.uk/gp/aw/d/019...nQL&ref=plSrch
Could you do one of them for an example please
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chocochip_
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Hi there,
The answer to question one is:
Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20
Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:
The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:
25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:
0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n
Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)
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Lucky10
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(Original post by noor.m)
Hi there,
The answer to question one is:
Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20
Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:
The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:
25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:
0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n
Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)
Could you just give me a breakdown using numbers how you did 2a and c
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Lucky10
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(Original post by noor.m)
Hi there,
The answer to question one is:
Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20
Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:
The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:
25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:
0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n
Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)
I thought the balanced equation for ethanol was C2H5OH(l) + 3 O2(g) =Heat=> 2 CO2(g) + 3 H2O(g)
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Lucky10
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(Original post by noor.m)
Hi there,
The answer to question one is:
Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20
Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:
The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:
25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:
0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n
Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)
And for propanol 2C3H7OH + 9 O2 >> 6 CO2 + 8H2O
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chocochip_
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(Original post by Lucky10)
And for propanol 2C3H7OH + 9 O2 >> 6 CO2 + 8H2O
I think there's two ways of balancing it?
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Lucky10
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(Original post by noor.m)
I think there's two ways of balancing it?
Might be, could you give me a breakdown using numbers with question 2a and c please
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Lucky10
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(Original post by noor.m)
I think there's two ways of balancing it?
Wb q4??
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chocochip_
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(Original post by Lucky10)
Wb q4??
Um i think it's because the data-book is only an average of the enthalpy change
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